Chemical Engineering Calculations - Dimensional Analysis Problems

CHEMICAL ENGINEERING CALCULATIONS
DIMENSIONAL ANALYSIS PROBLEMS



𝑐𝑐𝑐𝑐𝑐𝑐 𝐵𝐵𝐵𝐵𝐵𝐵
1. Convert 1 𝑡𝑡𝑡𝑡
𝑔𝑔−℃ 𝑙𝑙𝑏𝑏𝑚𝑚 −℉
𝑐𝑐𝑐𝑐𝑐𝑐
Given: 1
𝑔𝑔−℃
𝐵𝐵𝐵𝐵𝐵𝐵
Required: conversion to
𝑙𝑙𝑏𝑏𝑚𝑚 −℉
Solution:

𝑐𝑐𝑐𝑐𝑐𝑐 0.004 𝐵𝐵𝐵𝐵𝐵𝐵 1 𝑘𝑘𝑘𝑘 1000 𝑔𝑔 1 ℃ 𝐵𝐵𝐵𝐵𝐵𝐵
=1 � �� �� � =1
𝑔𝑔 − ℃ 1 𝑐𝑐𝑐𝑐𝑐𝑐 2.2 𝑙𝑙𝑏𝑏𝑚𝑚 1 𝑘𝑘𝑘𝑘 1.8 ℉ 𝑙𝑙𝑏𝑏𝑚𝑚 − ℉
𝑩𝑩𝑩𝑩𝑩𝑩
Answer: 𝟏𝟏
𝒍𝒍𝒃𝒃𝒎𝒎 −℉


2. At what temperature reading do the Fahrenheit scale and Centigrade scale coincide?
Given: ℉ 𝑎𝑎𝑎𝑎𝑎𝑎 ℃
Required: temperature reading where both will coincide
Solution:

9
℉ = (℃ + 32)
5
5
℃ = (℉ − 32)
9

Celsius:
9
℃ = (℃ + 32)
5
9
℃ − ℃ = 32
5
4
− ℃ = 32
5
℃ = −40

Fahrenheit
5
℉ = (℉ − 32)
9
5 160
℉= ℉−
9 9
5 160
℉− ℉=−
9 9
4 160
℉=−
9 9
℉ = −40

Answer: ∴ 𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 𝒂𝒂𝒂𝒂𝒂𝒂 𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭𝑭 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 𝒂𝒂𝒂𝒂 − 𝟒𝟒𝟒𝟒 𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅𝒅

3. Convert the following temperature readings
a. 180 K to ℃, °𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 ℉
Given: 180 K
Required: conversion to ℃, °𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 ℉
Solution:
To ℃: 180 𝐾𝐾 − 273.15 = −93.15℃
To °𝑅𝑅: (180 𝐾𝐾)(1.8) = 324°𝑅𝑅
9
To ℉: (180 𝐾𝐾 − 273.15) + 32 = −135.67℉
5

, b. 800 ℉ to ℃, °𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾
Given: 800 °F
Required: conversion to ℃, °𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 𝐾𝐾
Solution:
5
To ℃: (800°𝐹𝐹 − 32) = 426.67°𝐶𝐶
9
To °𝑅𝑅: 800°𝐹𝐹 + 459.67 = 1259.67°𝑅𝑅
5
To K: (800℉ − 32) + 273.15 = 699.82 K
9
Answer: 𝟔𝟔𝟔𝟔𝟔𝟔. 𝟖𝟖𝟖𝟖 𝐊𝐊, 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟔𝟔𝟔𝟔°𝑪𝑪, 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔°𝑹𝑹

c. 100 000 ℃ to 𝐾𝐾, °𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 ℉
Given: 100,000 °C
Required: conversion to 𝐾𝐾, °𝑅𝑅 𝑎𝑎𝑎𝑎𝑎𝑎 ℉
Solution:
To K: 100 000°𝐶𝐶 + 273.15 = 100 273.15 𝐾𝐾
9
To °𝑅𝑅: (100 000°𝐶𝐶) + 491.67 = 180 491.67°𝑅𝑅
5
9
To ℉: (100 000°𝐶𝐶) + 32 = 180 032°𝐹𝐹
5
Answer: 𝟏𝟏𝟏𝟏𝟏𝟏 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟏𝟏𝟏𝟏 𝑲𝑲, 𝟏𝟏𝟏𝟏𝟏𝟏 𝟎𝟎𝟎𝟎𝟎𝟎°𝑭𝑭, 𝟏𝟏𝟏𝟏𝟏𝟏 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟔𝟔𝟔𝟔°𝑹𝑹

𝐵𝐵𝐵𝐵𝐵𝐵 𝐽𝐽
4. Convert a thermal conductivity of 0.003 to
ℎ𝑟𝑟 −℉−𝑓𝑓𝑓𝑓 𝑠𝑠−𝐾𝐾−𝑘𝑘
𝐵𝐵𝐵𝐵𝐵𝐵
Given: 0.003
ℎ𝑟𝑟 −℉−𝑓𝑓𝑓𝑓
𝐽𝐽
Required: conversion to
𝑠𝑠−𝐾𝐾−𝑘𝑘
Solution:
𝑊𝑊
𝐵𝐵𝐵𝐵𝐵𝐵 1.7307
0.003 ∙ 𝐾𝐾 ∙ 𝑚𝑚 = 5.1921 𝑥𝑥 10−3 𝑊𝑊 𝑜𝑜𝑜𝑜 𝐽𝐽
ℎ𝑟𝑟. °𝐹𝐹. 𝑓𝑓𝑓𝑓 1 𝐵𝐵𝐵𝐵𝐵𝐵 𝐾𝐾 ∙ 𝑚𝑚 𝑠𝑠 − 𝐾𝐾 − 𝑘𝑘
ℎ𝑟𝑟. °𝐹𝐹. 𝑓𝑓𝑓𝑓
𝑱𝑱
Answer: 𝟓𝟓. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟑𝟑
𝒔𝒔−𝑲𝑲−𝒌𝒌


5. The emissive power of a blackbody depends on the fourth power of the temperature and is given
𝐵𝐵𝐵𝐵𝐵𝐵
by 𝑊𝑊 = 𝐴𝐴𝑇𝑇 4 ; where: W= emissive power � �; A= Stefan-Boltzmann constant,
𝑓𝑓𝑡𝑡 2 −ℎ𝑟𝑟
𝐵𝐵𝐵𝐵𝐵𝐵 𝐽𝐽
0.171 𝑥𝑥 10−8 ; T= temperature in °𝑅𝑅. What is the value of A in the units
𝑓𝑓𝑡𝑡 2 −ℎ𝑟𝑟−°𝑅𝑅4 𝑚𝑚2 −𝑠𝑠−𝐾𝐾 4
Given:
𝑊𝑊 = 𝐴𝐴𝑇𝑇 4
𝐵𝐵𝐵𝐵𝐵𝐵
where: W= emissive power � �
𝑓𝑓𝑡𝑡 2 −ℎ𝑟𝑟
𝐵𝐵𝐵𝐵𝐵𝐵
A= Stefan-Boltzmann constant �0.171 𝑥𝑥 10−8 2 �
𝑓𝑓𝑡𝑡 −ℎ𝑟𝑟−°𝑅𝑅4
T= temperature (°𝑅𝑅)
𝐽𝐽
Required: value of A in the units 2
𝑚𝑚 −𝑠𝑠−𝐾𝐾4
Solution:
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶: 1°𝑅𝑅 = 0.556 𝐾𝐾
1 𝑓𝑓𝑓𝑓 = 0.3014 𝑚𝑚
1 ℎ𝑟𝑟 = 3600 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
1 𝐵𝐵𝐵𝐵𝐵𝐵 = 1055.06 𝐽𝐽

𝐵𝐵𝐵𝐵𝐵𝐵 1055.06 𝐽𝐽 (1 𝑓𝑓𝑓𝑓)2 1 ℎ𝑟𝑟 (1 °𝑅𝑅)4
�0.171 𝑥𝑥 10−8 � �� � � � � � � ��
𝑓𝑓𝑡𝑡 2 − ℎ𝑟𝑟 − °𝑅𝑅 4 1 𝐵𝐵𝐵𝐵𝐵𝐵 (0.3014 𝑚𝑚)2 3600 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (0.556 𝐾𝐾)4
𝐽𝐽
= 5.7728 𝑥𝑥 10−8 2
𝑚𝑚 − 𝑠𝑠 − 𝐾𝐾 4

𝑱𝑱
Answer: 𝟓𝟓. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕 𝒙𝒙 𝟏𝟏𝟏𝟏−𝟖𝟖
𝒎𝒎𝟐𝟐 −𝒔𝒔−𝑲𝑲𝟒𝟒